An R-module M is called Matlis injective if Ext1R(E(R), M

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http://dx.doi.org/10.4134/BKMS.2013.50.2.459

MATLIS INJECTIVE MODULES

Hangyu Yan

Abstract. In this paper, Matlis injective modules are introduced and studied. It is shown that every R-module has a (special) Matlis injective preenvelope over any ring R and every right R-module has a Matlis injective envelope when R is a right Noetherian ring. Moreover, it is shown that every right R-module has an F^⊥¹-envelope when R is a right Noetherian ring and F is a class of injective right R-modules.

1. Introduction

Throughout this paper, R will denote an associative ring with identity and all modules will be unitary right R-modules.

The motivation of this paper is from [4], where the notion of Whitehead modules was studied. Recall that an R-module M is called a Whitehead module or W -module if Ext¹_R(M, R) = 0. We introduce the notion of Matlis injective modules as a dual notion of Whitehead modules in some sense. An R-module M is called Matlis injective if Ext¹_R(E(R), M ) = 0, where E(R) denotes the injective envelope of R. Let R be an integral domain and Q its field of quotients, an R-module C is called Matlis cotorsion or weakly cotorson if Ext¹_R(Q, C) = 0.

Then, it is easy to see that the notion of Matlis injective R-modules coincides with the notion of Matlis cotorsion R-modules when R is an integral domain.

Following [7], an R-module M is called copure injective if Ext¹_R(E, M ) = 0 for any injective R-module E. Clearly, every copure injective R-module is Matlis injective, but it is easy to see that the converse is not true in general. Thus Matlis injective R-modules can be seen as a generalization of copure injective R-modules.

Let C be a class of R-modules. Enochs defined a C-(pre)cover ( C-(pre)envelope) of an R-module in [6]. Therefore, it is natural to study the existence of Matlis injective (pre)covers and Matlis injective (pre)envelopes. Obviously, the class of Matlis injective R-modules is closed under direct summands, but we show that it is not closed under direct sums in general. So there exist a ring

Received October 15, 2011; Revised January 20, 2012.

2010 Mathematics Subject Classification. 16D10, 16E30.

Key words and phrases. Matlis injective module, (pre)envelope, Σ-pure injective.

This research was partially supported by Research Fund of China Pharmaceutical Uni- versity (No. 211148).

c

2013 The Korean Mathematical Society 459

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R and an R-module M such that M doesn’t have a Matlis injective precover.

Then, we are only interested in the existence of Matlis injective (pre)-envelopes in this paper. Let F be a class of R-modules, we denote by F^⊥¹ the class of R-modules N such that Ext¹_R(F, N ) = 0 for every F ∈ F . In [5, Theorem 10], Eklof and Trlifaj proved that if there is a set S of R-modules such that F^⊥¹ = S^⊥¹, then every R-module has an F^⊥¹-preenvelope. Using this result, we show that every R-module has a Matlis injective preenvelope. If R is a right Noetherian ring, we show that every R-module has an F^⊥¹-envelope, where F is any subclass of the class of injective R-modules. As a byproduct, we show that every R-module has a Matlis injective envelope when R is a right Noetherian ring.

2. Preliminaries

In this section we briefly recall some definitions and results required in this paper.

For a ring R, Mod-R will denote the category of all right R-modules and pd(M ) will denote the projective dimension of M. For an R-module M, we denote by E(M ) the injective envelope of M. We frequently identify M with its image in E(M ) and think of M as a submodule of E(M ).

Let C ⊆Mod-R. Define

C^⊥¹ = {X ∈ Mod-R | Ext¹_R(C, X) = 0 for all C ∈ C},

⊥₁

C = {X ∈ Mod-R | Ext¹_R(X, C) = 0 for all C ∈ C}.

Add(C)={X ∈ Mod-R | X is a direct summand of L

i∈ICi, where I is a set and where for any i ∈ I, Ci is isomorphic to an element of C}.

For C = {C}, we write C^⊥¹,^⊥¹C and Add(C) in place of {C}^⊥¹,^⊥¹{C} and Add({C}), respectively.

Let M ∈ Mod-R. A homomorphism f ∈ HomR(M, C) with C ∈ C is called a C-preenvelope of M provided that the abelian group homomorphism HomR(f, C^′) : HomR(C, C^′) → HomR(M, C^′) is surjective for each C^′ ∈ C. The C-preenvelope f is called a C-envelope of M provided that f = gf implies g is an automorphism for each g ∈ EndR(C). Moreover, a C-preenvelope f : M → C of M is called special provided that f is injective and Coker f ∈^⊥¹C. C-envelopes may not exist in general, but if they exist, they are unique up to isomorphism.

If C is the class of injective modules, then we get the usual injective envelopes.

C-precovers and C-covers are defined dually. These generalize the projective covers introduced by Bass in the 1960’s.

A pair (A, B) of R-module classes is called a cotorsion theory (or cotorsion pair) provided that A^⊥¹ = B and A =^⊥¹B. An R-module M is called cotorsion if Ext¹_R(F, M ) = 0 for any flat R-module F . Let F be the class of flat R-modules and C be the class of cotorsion R-modules, it is known that (F , C ) is a cotorsion theory.

For any class F of R-modules. The following theorem, due to Eklof and

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Trlifaj, says that every R-module has a special F^⊥¹-preenvelope if there is a set S of R-modules such that S^⊥¹ = F^⊥¹. Before stating the result, we need more notions:

A sequence of modules A = (Aα| α ≤ µ) is called a continuous chain of modules provided that A0 = 0, Aα⊆ Aα+1 for all α < µ and Aα =S

β<αAβ

for all limit ordinals α ≤ µ.

Let M be a module and C a class of modules. Then M is called C-filtered provided that there are an ordinal κ and a continuous chain, (Mα| α ≤ κ), consisting of submodules of M such that M = Mκ, and such that each of the modules Mα+1/Mα(α < κ) is isomorphic to an element of C. The chain (Mα| α ≤ κ) is called a C-filtration of M.

Theorem 2.1 ([10], Theorem 3.2.1, p. 117). Let S be a set of R-modules and M an R-module. Then there is a short exact sequence 0 → M ֒→ P → N → 0, where P ∈ S^⊥¹ andN is S-filtered. In particular, M ֒→ P is a special S^⊥¹- preenvelope of M.

The following theorem from [10] gives a criterion to judge when an R-module M has a C^⊥¹-envelope.

Theorem 2.2 ([10], Theorem 2.3.2, p. 107). Let R be a ring and M be an R-module. Let C be a class of R-modules closed under extensions and direct limits. Assume thatM has a special C^⊥¹-preenvelopeν with Coker ν ∈ C. Then M has a C^⊥¹-envelope.

A short exact sequence 0 → A → B → C → 0 of R-modules is called pure if the induced sequence 0 → HomR(F, A) → HomR(F, B) → HomR(F, C) → 0 of abelian groups is exact for every finitely presented R-module F. A submodule A of an R-module B is called a pure submodule of B if the canonical exact sequence 0 → A → B → B/A → 0 is pure. An R-module M is called pure in- jectiveif the sequence 0 → HomR(C, M ) → HomR(B, M ) → HomR(A, M ) → 0 is exact for every pure exact sequence 0 → A → B → C → 0 of R-modules.

Let M be an R-module. M is said to be Σ-pure injective if for every index set I the direct sum M^(I) is pure injective. M is said to be Σ-self orthogonal if Ext¹_R(M, M^(I)) = 0 for every index set I.

The following property of Σ-pure injective modules will be used in this paper.

Proposition 2.3 ([9], Corollary 1.42, p. 30). Every pure submodule of a Σ-pure injective moduleB is a direct summand of B.

For unexplained terminology and notation, we refer the reader to [1, 3, 8, 10, 13].

3. Properties of Matlis injective modules We start with the following definition.

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Definition 3.1. Let R be a ring and M an R-module. M is said to be Matlis injective if Ext¹_R(E(R), M ) = 0. An R-module N is said to be Matlis projective if Ext¹_R(E(R), C) = 0 implies Ext¹_R(N, C) = 0 for any R-module C. R is said to be a right Matlis ring if E(R) is flat and pd(E(R)) ≤ 1.

In what follows, we denote by MI (MP) the class of Matlis injective (projective) R-modules. For C = MI, C-(pre)envelopes will simply be called Matlis injective (pre)envelopes.

Proposition 3.2. Let R be a ring. Then MI is closed under extensions, direct products and direct summands; MI=Mod-R if and only if E(R) is projective.

Proof. It is easy to see that the assertion holds by definition. Corollary 3.3. Let R be an integral domain. Then every R-module is Matlis injective if and only if R is a field.

Proof. “ ⇐= ” is trivial.

“ =⇒ ”. By Proposition 3.2, E(R) is projective, then there exists a non- zero homomorphism f ∈ HomR(E(R), R). So f (E(R)) is a non-zero divisible submodule of R. Let r be any non-zero element from R. We choose a non-zero element x ∈ f (E(R)). Since rx is non-zero and f (E(R)) is divisible, there is an element y ∈ f (E(R)) with (rx)y = x, and so (ry − 1)x = 0. But R is an integral domain, then ry − 1 = 0, i.e., ry = 1. Hence R is a field. Remark 3.4. Recall that a commutative domain R is called almost perfect provided that R/I is a perfect ring for each ideal 0 6= I 6= R. We will show that MI is not closed under direct sums if R is an almost perfect domain but not a field. If R is an almost perfect domain, then MI coincides with the class of cotorsion R-modules by [10, Theorem 4.4.16, p. 172]. But the class of cotorsion R-modules is closed under direct sums if and only if R is a perfect ring by [11, Theorem 19]. Note that E(R) is flat when R is a commutative domain, and so R is a perfect ring if and only if R is a field by Corollary 3.3. Hence MI is not closed under direct sums when R is an almost perfect domain but not a field.

Then we will show that there exist a ring R and an R-module M such that M doesn’t have a Matlis injective precover. For example, let R be an almost perfect domain but not a field, then there exists a family {Mi}i∈I of Matlis injective R-modules such thatL

i∈IMi is not Matlis injective. But since MI is closed under direct summands by Proposition 3.2, it is easy to check that L

i∈IMi doesn’t have a Matlis injective precover.

Lemma 3.5. Let R be a ring. Then every cotorsion R-module is Matlis injective if and only if E(R) is flat.

Proof. “ ⇐= ” is clear.

“ =⇒ ”. Let C be any cotorsion R-module. By hypothesis, we have Ext¹_R(E(R), C) = 0.

Hence E(R) is flat by the fact that (F , C ) is a cotorsion theory.

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Proposition 3.6. Let R be a ring. Then MI = C if and only if E(R) is flat and every Matlis injective R-module is cotorsion.

Proof. “ ⇐= ” holds by assumption and Lemma 3.5.

“ =⇒ ”. By assumption, we have M is cotorsion if and only if it is Matlis injective. Then the assertion holds by Lemma 3.5. Proposition 3.7. Let R be a ring. Then the following are equivalent.

(1) Every quotient module of any Matlis injective R-module is Matlis injective.

(2) Every quotient module of any injective R-module is Matlis injective.

(3) The projective dimension of E(R) is at most 1.

Proof. (1) =⇒ (2) is trivial.

(2) =⇒ (3). Let K be any R-module. It is enough to show that Ext²_R(E(R), K) = 0. Let us consider the exact sequence 0 → K → E(K) → E(K)/K → 0.

We then have the exact sequence Ext¹_R(E(R), E(K)/K) → Ext²_R(E(R), K) → Ext²_R(E(R), E(K)) = 0. Note that Ext¹_R(E(R), E(K)/K) = 0 by (2), we get Ext²_R(E(R), K) = 0.

(3) =⇒ (1). Let M be a Matlis injective R-module and N a submodule of M. Let us consider the exact sequence 0 → N → M → M/N → 0. Applying the functor HomR(E(R), −) to the above exact sequence, we get the exact sequence 0 = Ext¹_R(E(R), M ) → Ext¹_R(E(R), M/N ) → Ext²_R(E(R), N ). Note that Ext²_R(E(R), N ) = 0 by (3), so Ext¹_R(E(R), M/N ) = 0 and (1) follows. Remark 3.8. If E(R) is flat, then the condition that every quotient module of any cotorsion R-module is Matlis injective is also equivalent to the conditions of Proposition 3.7.

Lemma 3.9. Let R be a ring. Then (MP, MI) is a cotorsion theory.

Proof. Straightforward.

Theorem 3.10. Let R be a ring. Then the following are equivalent.

(1) R is a right Matlis ring.

(2) Every quotient module of any Matlis injective R-module is Matlis injective and every cotorsionR-module is Matlis injective.

(3) Every quotient module of any injective R-module is Matlis injective and every cotorsionR-module is Matlis injective.

(4) Every Matlis projective R-module is flat and its projective dimension is at most1.

Proof. (1) ⇐⇒ (2) ⇐⇒ (3) hold by Lemma 3.5 and Proposition 3.7.

(1) =⇒ (4). By Lemma 3.9 and [10, Corollary 3.2.4, p. 119], every Matlis projective R-module is a direct summand of some {E(R), R}-filtered R-module.

Note that every {E(R), R}-filtered R-module is flat and its projective dimension is at most pd(E(R)) by (1) and [10, Lemma 3.1.2, p. 113]. So every Matlis projective R-module is flat and its projective dimension is at most 1.

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(4) =⇒ (1). Obviously, E(R) is Matlis projective by definition. So (1) holds

by assumption.

Recall that a submodule N of a module M of projective dimension k is said to be a tight submodule if the projective dimension of M/N is at most k. We now have the following simple fact:

Proposition 3.11. Let R be a ring. If pd(E(R)) ≤ 1, then tight submodules of Matlis projective R-modules are also Matlis projective.

Proof. Let us consider the exact sequence 0 → N → M → M/N → 0, where M is Matlis projective and N is a tight submodule of M. Then pd(M ) ≤ 1 by hypothesis and the proof of Theorem 3.10. For any Matlis injective R-module C, we have the induced exact sequence

Ext¹_R(M, C) → Ext¹_R(N, C) → Ext²_R(M/N, C).

The two ends vanish, since M is Matlis projective and pd(M/N ) ≤ pd(M ) ≤ 1.

So the middle term is 0, and hence the assertion holds. Proposition 3.12. Let R be a ring. Then the following are equivalent.

(1) C ∈ MI whenever 0 → A → B → C → 0 is an exact sequence of R-modules such that A, B ∈ MI.

(2) E(M )/M is Matlis injective when M is Matlis injective.

(3) For any R-module M, Ext¹_R(E(R), M ) = 0 implies Ext²_R(E(R), M ) = 0.

Proof. (1) =⇒ (2) is trivial.

(2) =⇒ (3). Let M be an R-module such that Ext¹_R(E(R), M ) = 0, i.e., M is Matlis injective. Then E(M )/M is Matlis injective by (2). Applying the functor HomR(E(R), −) to the exact sequence 0 → M → E(M ) → E(M )/M → 0, we have the exact sequence 0 = Ext¹_R(E(R), E(M )/M ) → Ext²_R(E(R), M ) → Ext²_R(E(R), E(M )) = 0. So Ext²_R(E(R), M ) = 0.

(3) =⇒ (1). Let 0 → A → B → C → 0 be an exact sequence of R-modules such that A, B ∈ MI. Applying the functor HomR(E(R), −) to the above sequence, we have the exact sequence 0 = Ext¹_R(E(R), B) → Ext¹_R(E(R), C) → Ext²_R(E(R), A) = 0 by (3). So Ext¹_R(E(R), C) = 0, i.e., C is Matlis injective.

Hence (1) holds.

Proposition 3.13. Let R be a commutative Artinian ring. Then MI is closed under direct sums, pure submodules and direct limits. Moreover, MI is a definable class, i.e., it is closed under pure submodules, direct products and direct limits.

Proof. By hypothesis, E(R) is finitely presented by [12, Theorem 3.64, p. 90].

Then MI is closed under direct sums by the isomorphism MExt¹_R(F, Mα) ∼= Ext¹R(F,M

Mα)

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for any finitely presented R-module F and any family {Mα} of R-modules.

Suppose that A is a pure submodule of a Matlis injective R-module B. Then we have the exact sequences 0 −→ HomR(E(R), A) −→ HomR(E(R), B) −→

HomR(E(R), B/A) −→ 0 and HomR(E(R), B) −→ HomR(E(R), B/A) −→

Ext¹_R(E(R), A) −→ Ext¹_R(E(R), B) = 0. Hence Ext¹_R(E(R), A) = 0, i.e., A is Matlis injective. So MI is closed under pure submodules. That MI is closed under direct limits follows from the isomorphism Ext¹_R(F, lim−→Mⁱ) ∼= lim−→ Ext¹^R(F, Mi) for any finitely presented R-module F and any family {Mi} of R-modules since R is a commutative Artinian ring. So MI is definable by

Proposition 3.2.

Proposition 3.14. Let R be a commutative Artinian ring and S ⊂ R be a multiplicative set. If M is a Matlis injective R-module, then S⁻¹M is a Matlis injective S⁻¹R-module.

Proof. By assumption, E(R) is finitely generated by [12, Theorem 3.64, p. 90]

and R is a Noetherian ring. So,

Ext¹_S−1R(S⁻¹ER(R), S⁻¹M ) ∼= S⁻¹Ext¹R(ER(R), M )

by [8, Theorem 3.2.5, p. 76]. But S⁻¹ER(R) ∼= ES⁻¹R(S⁻¹R) by [8, Theorem 3.3.3, p. 84]. Thus S⁻¹M is a Matlis injective S⁻¹R-module when M is a

Matlis injective R-module.

Proposition 3.15. Let R be a commutative Noetherian ring and S ⊂ R be a multiplicative set. If M is a Matlis projective R-module, then S⁻¹M is a Matlis projective S⁻¹R-module.

Proof. Note that S⁻¹ER(R) ∼= ES⁻¹R(S⁻¹R) by [8, Theorem 3.3.3, p. 84]

and by hypothesis. Then, every Matlis projective S⁻¹R-module is a direct summand of some {S⁻¹ER(R), S⁻¹R}-filtered S⁻¹R-module by [10, Corollary 3.2.4, p. 119]. Since M is a Matlis projective R-module, M is a direct summand of some {E(R), R}-filtered R-module by [10, Corollary 3.2.4, p. 119]. Let N be an {E(R), R}-filtered R-module and the chain (Nα| α ≤ κ) be a {E(R), R}- filtration of N . Then S⁻¹N is an {S⁻¹ER(R), S⁻¹R}-filtered S⁻¹R-module and the chain (S⁻¹Nα| α ≤ κ) is a {S⁻¹ER(R), S⁻¹R}-filtration of S⁻¹N by [8, Theorem 1.5.7, p. 33, and Proposition 2.2.4, p. 44] and by definition. So S⁻¹M is a Matlis projective S⁻¹R-module and the assertion holds.

4. The existence of Matlis injective (pre)envelopes

According to Theorem 2.1, we immediately have the following proposition.

Proposition 4.1. Let R be a ring. Then every R-module has a special Matlis injective preenvelope.

The following lemmas are needed to prove the main result of this paper.

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Lemma 4.2. Let R be a ring and M an R-module. If M is Σ-pure injective and Σ-self orthogonal, then Add(M ) is closed under extensions and direct limits.

Proof. Let 0 → A → B → C → 0 be an exact sequence of R-modules such that both A and C are in Add(M ). Without loss of generality, we may assume that both A and C are direct summands of M^(I) for an index set I. Since M is Σ-self orthogonal, we have Ext¹_R(C, A) = 0. Then the exact sequence 0 → A → B → C → 0 splits, and so B ∼= AL

C. Obviously, AL

C ∈ Add(M ).

Therefore, B ∈ Add(M ). So Add(M ) is closed under extensions. We claim that any R-module N from Add(M ) is Σ-pure injective. It is clear that N is pure injective since M is Σ-pure injective. In addition, Add(M ) is closed under direct sums. Thus N is Σ-pure injective. Let ((Mi)i∈I, (fji)) be a direct system of R-modules from Add(M ) where I is a directed set. Then there exists a short exact sequence 0 → K ֒→ L

i∈IMi → lim−→Mⁱ → 0 with K a pure submodule ofL

i∈IMi. ButL

i∈IMi is Σ-pure injective, then the exact sequence 0 → K ֒→ L

i∈IMi → lim−→Mⁱ → 0 splits by Proposition 2.3. So lim−→Mⁱ is isomorphic to a direct summand ofL

i∈IMi, i.e., lim−→Mⁱ ∈ Add(M ).

Hence Add(M ) is closed under direct limits.

Lemma 4.3. Let R be a ring and M an R-module. Assume that M is Σ-pure injective andΣ-self orthogonal. Then every R-module N has an M^⊥¹-envelope.

Proof. Obviously, M^⊥¹ = (Add(M ))^⊥¹. Thus it is equivalent to show that every R-module N has an (Add(M ))^⊥¹-envelope. By Theorem 2.1, N has a special (Add(M ))^⊥¹-preenvelope f with Coker f is {M }-filtered. Note that every {M }-filtered R-module is in Add(M ) by Lemma 4.2 and transfinite induction.

So N has an (Add(M ))^⊥¹-envelope by Lemma 4.2 and Theorem 2.2. We are now in a position to prove the following

Theorem 4.4. Let R be a right Noetherian ring and F a class of injective R- modules. Then every R-module M has an F^⊥¹-envelope; in particular, every R-module M has a Matlis injective envelope.

Proof. If R is right Noetherian, then every injective R-module is the direct sum of indecomposable injective R-modules. Each such module is the injective envelope of a cyclic R-module. Hence, we can find a representative set of such modules. So there is a family {Ei}i∈I of indecomposable injective R-modules such that every injective R-module is the direct sum of copies of Ei.

Let S = {Ei| Eiis isomorphic to a direct summand of an element of F }. It is easy to see that (L

Ei∈SEi)^⊥¹ = F^⊥¹. Note thatL

Ei∈SEi is Σ-pure injective and Σ-self orthogonal by the fact that the class of right injective R-modules is closed under direct sums when R is right Noetherian. So the assertion holds

by Lemma 4.3.

We end this paper with the following remark.

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Remark 4.5. If R is a commutative Artinian ring, then every R-module has a Matlis injective cover by Proposition 3.13 and [2, Corollary 2.6 and Proposition 4.3(3)].

Acknowledgements. The author would like to thank the referees for the very helpful comments and suggestions.

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Faculty of Science

China Pharmaceutical University Nanjing 211198, P. R. China E-mail address: [email protected]